CBSE Class 12 Biology Question Paper 2024 (Set 2- 57/1/2) Available- Download Solution PDF with Answer Key

Solution: In a polynucleotide chain, the dotted line (‘—') represents the bond between the nitrogenous base (B) and the sugar (S). This bond is known as an N-glycosidic linkage. It connects the 1' carbon of the pentose sugar to the nitrogen atom of the nitrogenous base.

Other bonds in the polynucleotide structure: Phosphodiester bond: Links the sugar molecules of adjacent nucleotides. Hydrogen bond: Occurs between complementary nitrogenous bases in double-stranded DNA. Peptide bond: Found in proteins, not in polynucleotides.

Solution: The human female reproductive cycle consists of several stages regulated by hormones. The correct sequence of events involves the following steps:

1. Secretion of FSH (Follicle-Stimulating Hormone): The cycle begins with the secretion of FSH from the anterior pituitary gland. FSH stimulates the growth and maturation of the ovarian follicles in the ovaries, which is the first step in the process.

2. Growth of follicle and oogenesis: As FSH is secreted, the ovarian follicles start to grow and develop. This is accompanied by the process of oogenesis, where the primary oocyte completes its maturation into a secondary oocyte within the growing follicle.

3. Sudden increase in LH (Luteinizing Hormone): Just before ovulation, there is a surge in LH, which is triggered by high levels of estrogen produced by the growing follicle. This LH surge is the key trigger for ovulation.

4. Ovulation: The sudden rise in LH causes the mature follicle to rupture and release the secondary oocyte. This process is called ovulation, which occurs roughly in the middle of the menstrual cycle.

5. Growth of corpus luteum: After ovulation, the ruptured follicle transforms into the corpus luteum. This structure secretes progesterone, which prepares the uterine lining for a potential pregnancy.

The correct sequence of events is as follows: Secretion of FSH (i) - Growth of follicle and oogenesis (iv) - Sudden increase in LH (v) - Ovulation (ii) - Growth of corpus luteum (iii)

Thus, the correct sequence is i → iv → v → ii → iii, which corresponds to option (B).

Solution:
• Lactobacillus is a heterotrophic bacterium that ferments lactose in milk to form lactic acid.
• This process causes milk to coagulate and turn into curd.
• Heterotrophic bacteria rely on organic substances for energy.

Solution:
• Bt toxin genes (Cry-I Ac and Cry-II Ab) are derived from the bacterium Bacillus thuringiensis.
• These genes are inserted into cotton plants to produce insecticidal proteins that protect against bollworms.

Solution: To determine the DNA sequence that codes for the given polypeptide sequence, follow these steps:

• mRNA codon for Alanine is GCA, so the DNA coding strand must contain CGT.
• mRNA codon for Arginine is CGA, which corresponds to DNA codon GCT.
• mRNA codon for Lysine is AAG, which corresponds to DNA codon TTC.
• mRNA codon for Phenylalanine is UUU, which corresponds to DNA codon AAA.

Thus, the correct DNA sequence is CGT GCT TTC AAA.

Solution:
• Maize is a monoecious plant where both male and female flowers are present on the same plant.
• Geitonogamy occurs when pollination happens between flowers on the same plant, while xenogamy occurs between flowers on different plants.
• Papaya and Date Palm are dioecious plants, and Spinach is not relevant in this context.

Solution:
• Transgenic mice have been genetically modified to carry human genes that make them susceptible to polio infection.
• They are used to test the safety and effectiveness of the polio vaccine before human trials.

Solution:
• Mast cells play a crucial role in allergic reactions by releasing histamines when exposed to allergens.
• Histamines cause symptoms like inflammation, itching, and increased vascular permeability.

Solution:
• Periodic abstinence involves avoiding intercourse during the fertile window.
• In a 28-day cycle, ovulation usually occurs around day 14. The fertile window is days 10 to 17 when conception chances are high.

Solution:
• The karyotype 45 + XX does not exist. The correct condition for a broad palm with characteristic palm crease is 47 + XXY (Klinefelter syndrome) or other related conditions.

Solution: In Polymerase Chain Reaction (PCR), the DNA amplification follows a process of exponential growth. In each cycle, the amount of DNA doubles. Therefore, after each cycle, the number of DNA molecules increases by a factor of 2.

If we start with one DNA molecule, the number of DNA molecules after n cycles of amplification can be calculated using the formula:

Number of DNA molecules = 2ⁿ

where n is the number of cycles.

Given that the question asks about 10 cycles:

Number of DNA molecules after 10 cycles = 2¹⁰ = 1024

Therefore, after 10 cycles of PCR amplification, the number of DNA molecules generated will be 1024.

Option (A): 512 is incorrect. It represents the number of DNA molecules after 9 cycles (2⁹ = 512). Option (B): 1024 is the correct answer. After 10 cycles, 2¹⁰ = 1024 DNA molecules are generated. Option (C): 2048 is incorrect. It represents the number of DNA molecules after 11 cycles (2¹¹ = 2048). Option (D): 256 is incorrect. It represents the number of DNA molecules after 8 cycles (2⁸ = 256).

Solution:
• Statement (i) is correct because both the lion (a carnivore) and the sparrow (a herbivore) are consumers in the ecological food chain.
• Statement (ii) is correct as the starfish Pisaster acts as a keystone predator, preventing the dominance of certain species and thus maintaining species diversity.
• Statement (iii) is incorrect because predators usually regulate prey populations rather than leading to their extinction.
• Statement (iv) is incorrect as the production of secondary metabolites like nicotine and strychnine in plants are defensive mechanisms, not metabolic disorders.

Solution: Agrobacterium tumefaciens is primarily a pathogen of dicot plants, not monocots. However, the statement about T-DNA is correct, as this bacterium transfers T-DNA into plant cells, causing crown gall disease.

Solution: Legal adoption is indeed permitted by Indian law and encouraged for childless couples. However, emotional and social factors are not absolute deterrents to adoption, as awareness and government initiatives have addressed these concerns.

Solution: Linked genes are present on the same chromosome and tend to be inherited together. As a result, they do not assort independently according to Mendel's law of independent assortment, and thus, the expected dihybrid ratio of 9:3:3:1 is not observed.

Solution: Adenosine deaminase (ADA) deficiency is treated by introducing functional ADA genes into the lymphocytes of the patient. However, this does not necessarily have to be done at the early embryonic stage; it can also be done later.

The statement that embryonic cells are immortal, differentiated, and easy to manipulate is incorrect. Embryonic cells are actually undifferentiated and have the potential to develop into various cell types, but they are not necessarily easy to manipulate.

Solution: Apomixis can be compared to asexual reproduction because it leads to the formation of offspring without fertilization. In apomixis, seeds are produced without the fusion of gametes, resulting in progeny that are genetically identical to the parent plant.

Benefits of apomixis to farmers:
• Ensures the production of hybrid seeds without segregation of traits.
• Reduces the cost of hybrid seed production.
• Provides disease-resistant and high-yielding crops.

Solution:
1. The doctor injected antivenom to neutralize the snake venom and prevent its harmful effects.
2. The immunity provided is passive immunity, as pre-formed antibodies are administered.

Solution: Organic farmers do not recommend complete eradication of insect pests because:
• Insects play a vital role in maintaining ecological balance.
• Some insects act as natural predators and control pest populations naturally.
• Eradication can disrupt the food chain and lead to the emergence of more resistant pest species.

Solution: The naturalist who provided the explanation for this example is Jean-Baptiste Lamarck.

According to Lamarck's theory of evolution, the long necks of giraffes evolved over generations due to the following explanation:
• Giraffes originally had shorter necks and fed on lower vegetation.
• Due to competition and environmental changes, they stretched their necks to reach higher leaves.
• This continuous use of neck muscles led to gradual elongation over generations.
• The acquired trait of a longer neck was passed on to the offspring.

This idea is known as the Theory of Inheritance of Acquired Characteristics.

Solution: The basic steps followed in the Assisted Reproductive Technologies (ART) programme are:
• Step 1: Hormonal stimulation is provided to the female to induce the production of multiple ova.
• Step 2: The eggs (ova) are retrieved from the ovaries and fertilized with sperm in a laboratory setting.
• Step 3: The fertilized egg (zygote) undergoes early embryonic development in vitro (outside the body).
• Step 4: The resulting embryo is then implanted into the uterus for further development.

It is known as the test tube baby programme because fertilization and early embryonic development occur outside the female body in a laboratory setting, commonly in a petri dish or test tube.

Solution:
(a) The plant introduced into Australia was Opuntia (commonly known as Prickly Pear Cactus). Its uncontrollable growth was due to the absence of natural predators or pathogens in the Australian ecosystem, leading to its rapid and unchecked spread.
(b) The spread of Opuntia was eventually controlled by introducing a biological control agent, the Cactoblastis cactorum moth, whose larvae feed on the cactus and significantly reduced its population.

Solution: Transgenic animals are those that have had foreign genes deliberately inserted into their genome. These genes are introduced to study gene functions, improve livestock traits, or produce biologically important substances.

Solution: The first transgenic cow is Rosie. Importance: Rosie produced milk enriched with the human protein alpha-lactalbumin, making it nutritionally more balanced for infants.

Solution: EcoRI: 1. "E" = Genus (Escherichia). 2. "co" = Species (coli). 3. "R" = Strain (RY13). 4. "I" = First identified enzyme from this strain.

Solution:

5'-GAATTC-3' (Cleavage Site)
EcoRI cuts between G/AATTC, resulting in sticky ends.

Solution:
• Pneumonia symptoms: A includes breathing difficulty, cough, and chest pain.
• Typhoid causative organism: B is Salmonella typhi.
• Common cold symptoms: C include nasal congestion, sore throat, and cough.
• Ringworm causative organism: D is Trichophyton.
• Ascariasis symptoms: E include internal bleeding, muscular pain, and anemia.
• F: Disease caused by Entamoeba histolytica is Amoebiasis.

Solution:
Mode of Action:
1. Contraceptive pills contain synthetic hormones (estrogen and progesterone) that prevent ovulation by inhibiting the secretion of FSH and LH from the pituitary gland.
2. They thicken cervical mucus, making it difficult for sperm to enter the uterus.
3. They alter the endometrial lining, preventing implantation of a fertilized egg.

Schedule for Effective Outcome:
Pills should be taken daily for 21 days starting from the 5th day of the menstrual cycle, followed by a 7-day gap during which withdrawal bleeding occurs.

Solution:
Type of DNA: Short Tandem Repeats (STRs) or Variable Number Tandem Repeats (VNTRs).

Characteristics:
1. They are non-coding, repetitive sequences of DNA.
2. They show high polymorphism, making them unique to each individual (except identical twins).

Solution:
1. Forensic Science: Used for identification in criminal investigations.
2. Paternity Testing: Determines biological relationships.

Solution: The blood smear is indicative of Sickle Cell Anemia. It is caused by a mutation in the gene that codes for the beta chain of hemoglobin, leading to the production of abnormal hemoglobin S (HbS).

Symptoms:
• Severe anemia and fatigue
• Pain episodes (sickle cell crisis)
• Swelling in hands and feet
• Frequent infections
• Delayed growth and vision problems

Solution: Sickle cell anemia follows an autosomal recessive inheritance pattern. Individuals with one defective allele (HbA/HbS) are carriers, while those with two defective alleles (HbS/HbS) suffer from the disease.

Genetic cross:
Parent 1 (Carrier): HbA/HbS X Parent 2 (Carrier): HbA/HbS
Gametes: HbA, HbS and HbA, HbS
Offspring: HbA/HbA (Normal), HbA/HbS (Carrier), HbS/HbS (Affected)

The ratio of offspring will be:
• 25% Normal (HbA/HbA)
• 50% Carriers (HbA/HbS)
• 25% Affected (HbS/HbS)

Solution:
Significance of Miller's Experiment:
• Demonstrated that organic molecules like amino acids could be synthesized abiotically under conditions resembling early Earth's atmosphere.
• Supported the theory that life originated from simple organic compounds.

Scientists: The experiment was based on the hypothesis of Alexander Oparin and J.B.S. Haldane.

Solution: Meteorite analysis revealed the presence of simple organic compounds, such as amino acids, on extraterrestrial objects. This suggests that organic molecules can form naturally, supporting the idea of abiotic synthesis.

Solution: Primary productivity refers to the rate at which organic matter is produced by plants (producers) through photosynthesis in an ecosystem. It is expressed in terms of energy or biomass.

Units: Primary productivity is measured in kcal/m²/year (energy) or g/m²/year (biomass).

Solution: The primary productivity in Rajasthan is lower than in Kerala due to differences in climatic and environmental conditions:
• Rajasthan: Being a desert region, it has limited water availability, high temperatures, and low vegetation cover, all of which reduce primary productivity.
• Kerala: It has a tropical climate with abundant rainfall, fertile soil, and dense vegetation, which enhance primary productivity.

Solution:
(a) The transformed colony is the one that grows on Plate M (with ampicillin) but does not grow on Plate N (with both ampicillin and tetracycline). Justification: The recombinant DNA disrupts the tetracycline resistance gene, making the bacteria sensitive to tetracycline, but retains ampicillin resistance, allowing growth on Plate M.
(b) The sites are called selectable marker sites. Role: These genes help in identifying and selecting transformed cells: Ampicillin resistance gene ensures growth on ampicillin-containing media. Tetracycline resistance gene helps distinguish recombinant bacteria (tetracycline-sensitive) from non-recombinant ones.
(c) 1. Restriction endonucleases: Cut DNA at specific sequences. 2. DNA ligase: Joins DNA fragments to form recombinant DNA.

OR

(c) Insertional inactivation of the β-galactosidase gene prevents the production of the enzyme, resulting in white colonies (instead of blue) on media with X-gal. This helps identify recombinant bacteria.

Solution:
(a) Region III has maximum species richness due to: 1. Favourable climatic conditions such as tropical climate with high rainfall and temperature. 2. High resource availability and ecological niches supporting a large number of species.
(b) Region I has minimum species richness because: 1. It likely represents a cold or arid climate, which is less suitable for supporting diverse life forms. 2. Limited resource availability and harsh environmental conditions restrict the number of species.
(c) This uneven distribution of species diversity is referred to as Beta Diversity. It reflects the variation in species composition between different ecosystems or regions.
(d) 1. Prokaryotes lack morphological diversity, making species identification challenging. 2. Many prokaryotic species cannot be cultured in laboratories, limiting data collection. 3. Molecular techniques like DNA sequencing are required to identify them, which is time-consuming and resource-intensive.

Solution: The property present in normal cells but lost in cancer cells is contact inhibition.

Contact inhibition refers to the process where normal cells stop dividing when they come into contact with neighboring cells. This mechanism ensures that cells grow in an organized manner and maintain proper tissue structure. It prevents overgrowth and helps to regulate tissue size. In cancer cells, this property is lost, and they continue to divide uncontrollably, even when they come into contact with other cells. This uncontrolled division results in tumor formation and cancer progression.

Solution: The genes that may become cancerous under certain conditions are oncogenes and tumor suppressor genes.

1. Oncogenes: Oncogenes are mutated forms of normal genes called proto-oncogenes. Proto-oncogenes normally promote cell growth and division. However, when mutated or overexpressed, they become oncogenes, leading to uncontrolled cell division and potentially causing cancer. Example: The Ras gene, when mutated, continuously signals the cell to divide, even when not needed, leading to uncontrolled cell proliferation.

2. Tumor Suppressor Genes: Tumor suppressor genes are responsible for inhibiting cell growth and promoting the repair of damaged DNA or apoptosis (programmed cell death). If these genes are mutated or inactivated, cells can bypass growth control mechanisms and continue to divide. Example: The p53 gene is a critical tumor suppressor gene. When it is mutated, it fails to prevent the division of cells with damaged DNA, which can lead to cancer development.

Solution:
1. Biopsy and Histopathology: A biopsy involves removing a small sample of tissue from a suspected tumor or abnormal growth. The tissue is then examined under a microscope through histopathology. The role of biopsy and histopathology is essential for diagnosing cancer. They help identify the presence of cancerous cells, determine the type of cancer, and assess its stage and grade. This also provides critical information on the malignancy of the tumor and helps plan appropriate treatment options.

2. Magnetic Resonance Imaging (MRI): MRI is a non-invasive imaging technique that uses magnetic fields and radio waves to produce detailed images of the inside of the body. It is especially useful in detecting soft tissue abnormalities such as brain, spinal cord, and breast cancers. The role of MRI in cancer diagnosis includes identifying and locating tumors, determining their size, and assessing whether cancer has spread to other parts of the body (metastasis). It is also used to monitor treatment progress and plan surgeries or radiation therapy.

Solution:
(i) 1. Primary effluent is passed into large aeration tanks to increase the oxygen supply to the effluent. This helps in the growth of aerobic microorganisms that break down the organic matter in the sewage, converting it into simpler substances. The process of aeration encourages the growth of these microorganisms, which helps in reducing the biochemical oxygen demand (BOD) of the water, making it less polluting.

2. The sediment formed is called sludge. It consists of organic and inorganic matter that settles at the bottom of the settling tank. The significance of sludge is that it contains a high concentration of organic matter, which can be further processed to extract biogas or treated for use as fertilizer. Proper disposal of sludge is crucial for preventing environmental pollution.

3. The final step involves the treatment of the effluent in the settling tank, where the remaining suspended particles, including microorganisms, are allowed to settle down. This is followed by the removal of the clear water (supernatant), which is relatively free from pollutants. This treated water is then released into water bodies. Further treatment may be performed, depending on the quality of the effluent.

(ii) • Rhizobium (Kingdom: Bacteria): Rhizobium is a nitrogen-fixing bacterium that forms a symbiotic relationship with leguminous plants. It helps in fixing atmospheric nitrogen into a form that the plants can use for growth, thereby acting as a natural fertilizer.

• Azotobacter (Kingdom: Bacteria): Azotobacter is a free-living nitrogen-fixing bacterium found in soil. It helps in fixing nitrogen from the atmosphere and making it available to plants, improving soil fertility. Unlike Rhizobium, Azotobacter does not form symbiotic relationships with plants but independently fixes nitrogen.

Solution: Four patterns of inheritance that deviate from Mendelian genetics are:

• Incomplete dominance: Neither allele is completely dominant, resulting in an intermediate phenotype. Example: In snapdragons, red (RR) and white (rr) flowers produce pink (Rr) flowers in the heterozygous condition.

• Codominance: Both alleles are equally expressed in the heterozygous condition. Example: Blood group AB in humans, where both I^A and I^B alleles are expressed.

• Multiple alleles: A gene has more than two alleles. Example: ABO blood group in humans, controlled by three alleles (I^A, I^B, i).

• Polygenic inheritance: Traits are controlled by multiple genes. Example: Human skin color, which is determined by several gene pairs.

Explanation:

• Incomplete dominance: The pink color in snapdragons arises as the red allele cannot completely mask the effect of the white allele.

• Codominance: In blood group AB, both I^A and I^B alleles are expressed, showing traits of both A and B blood groups.

• Multiple alleles: The ABO blood group system shows that more than two alleles can govern a trait, where I^A and I^B are codominant, and i is recessive.

Solution: The transcription unit includes:
• Promoter: Located upstream of the transcription start site.
• Coding region: Contains the sequence to be transcribed.
• Terminator: Signals the end of transcription.
5' Promoter - Coding Region (5' to 3') - Terminator 3'

Solution: The coding strand does not become the template strand. Instead, the complementary strand (non-coding strand) acts as the template strand during transcription.

Solution: Transcription produces a single-stranded RNA because:
• RNA polymerase synthesizes RNA complementary to only one strand (the template strand).
• The resulting RNA strand does not pair with the DNA template, ensuring a single-stranded molecule for translation.

Solution: The development of male gametophyte in an angiosperm occurs as follows:
• The pollen grain (microspore) represents the male gametophyte.
• Inside the pollen grain, the nucleus divides mitotically to form two cells:
    – A larger vegetative cell (tube cell) that controls pollen tube growth.
    – A smaller generative cell that divides mitotically to form two male gametes.
• At the three-celled stage, the pollen grain is ready for fertilization.

Solution:

Solution:

Solution: Primary follicles are formed during the fetal stage in the ovary of a developing female fetus.

Solution: The events include:
• Ovulation:
– Triggered by a surge in luteinizing hormone (LH) around the 14th day of the menstrual cycle.
– The mature Graafian follicle ruptures, releasing the ovum.
• Post-ovulation:
– The ruptured follicle transforms into the corpus luteum, secreting progesterone.
– Progesterone maintains the uterine lining for potential implantation.
• If fertilization does not occur:
– The corpus luteum degenerates, leading to a drop in progesterone levels.
– The uterine lining is shed, resulting in menstruation.